Momentum and Impulse

Transcription

1 analyse impulse, and momentum transfer, in collisions between objects moving along a straight line; Momentum The momentum (p) of a body is the product of its mass and velocity. P = mv. The unit is kilogram metre per second. Newton's Laws of motion 1. Every object remains at rest, or in uniform motion in a straight line, unless acted on by some external force.. The rate of change of momentum is equal to the resultant force causing the change. 3. Action and reaction are equal and opposite. Deductions from Newtons second law. Consider a body of mass 'm' changing its velocity from 'u' to 'v' in time 't' under the action of a constant force F. From Newton's second law of motion, mv - mu v - u F = since a = F.t = mv - mu t t The product of a constant force and the time for which it acts is called the impulse (I) of the force. I = F.t The unit is the newton second. (Ns) Impulse is the change in momentum i.e. I = p - p1. Thus the impulse can be measured by the change in momentum produced. Impulse and momentum are vectors. So whenever a force acts, the direction of all the following is the same: F, a, v, F. t, p. Notes on problem solving 1. As momentum is a vector, a sign convention in problems is essential.. The negative sign for the change in momentum indicates a loss of momentum. 3. Remember that F is the resultant forward force. Graphically - constant or non-constant forces. Since F. t = m v for a constant force, it follows that the impulse will always be given by the area under the force-time graph. This area also measures the change in momentum. Area under "F - t" graph = Impulse = momentum. Page 1

2 Notes 1. Calculations by either formula or graph involve v and v. In many cases the body starts from rest and then, and only then, does v equal the actual velocity, v.. If asked for "p" look for impulse, if asked for impulse look for "p". 3. If answering for impulse, the units are "Ns" if answering for "p", the units are "kg m/s". Conservation of momentum When A and B collide, the action of A on B is equal and opposite to that of B on A. (Newtons 3 rd ) Hence the rate of change of momentum of A is equal and opposite to the rate of change of momentum of B. A B Since the time of contact is the same for both, then the change in momentum of A is equal and opposite to the change in momentum of B. That is, THE TOTAL MOMENTUM BEFORE IMPACT EQUALS THE TOTAL MOMENTUM AFTER IMPACT. This is known as the law of conservation of momentum. P(total) is constant before, during and after the collision. Notes: 1. Remember that a sign convention is essential.. If the bodies collide and stay together, then the momentum after the collision pfinal = pinitial 3. mvfinal = mvinitial 4. Mathematically, problems on 'collision' or 'explosion' are similar, except that for an explosion, the momentum of the system before the blast is often zero. 5. p(total before the collision) = p(total after the collision) 6. Always draw a diagram 7. Any unit may be used for mass or velocity, as long as such units are consistent within the equation. Momentum and Kinetic Energy Since p = mv and KE = 1 mv then p = useful later on with problems with matter waves. Momentum transfer involving the Earth mke, or KE = p. This equation is also very m 1. Body rises under gravity - slows down and loses momentum to the earth.. Body falling under gravity - speeds up giving the earth equal and opposite momentum change. 3. Falling body hits the ground - its p is transferred to the earth. 4. Body slowed due to friction - gives the earth and equal and opposite p. 5. Body accelerated due to friction - gives the earth an equal and opposite p. Page

3 - PAST VCAA EXAM Examples 00 The car traveling at 11.0ms -1 collides head-on with a concrete pillar. The car comes to rest in a time of 0.10 s. The car comes to rest against the pillar. The mass of the car and occupants is 1.30 tonne. Question 7 Determine the average force on the car during the impact with the pillar. a = v/t = 11/0.1 = 110 ms - F = ma =1.30 x 10 3 x 110 = 1.43 x 10 5 N [actually 1.4 x 10 5 N due to sig. figs] = 143 kn Application of the impulse-momentum equation F.t =. p resulted in an average force of 1.4 x 10 5 N. Many students did well on this question even those who struggled on previous questions. Clearly, students are quite confident about applying the impulse-momentum equation for a simple collision. Question 8 Explain how the crumple zone of the car can minimise the extent of injuries experienced by the occupants of the car. (Assume that the occupants are wearing seatbelts.) Crumple Zone enables passenger cell to travel a greater distance while stopping, in a longer time. The change in momentum [p f p o ] is the same regardless of the time taken. Longer time allows lower average force [same impulse force x time] The explanation of how the crumple zone can minimise the extent of injuries experienced by the occupants could have been addressed by either an impulse-momentum or a work-energy approach. Students needed to address the following points in order to score marks: the crumple zone extends either the time or distance of the collision the change in momentum or the change in kinetic energy is a fixed quantity for the collision. Each quantity depends only on the initial and final velocities longer collision time/distance results in smaller force on the occupants and hence minimises the extent of injuries they may experience. Students were generally clear about the fact that the crumple zone increased the time or distance for the collision and this resulted in a lower force on the occupants. However, many students were unable to describe the fact that the change in momentum or kinetic energy was a constant and how this was necessary for an understanding of the relationship between force and time or distance. Page 3

4 001 Jack and Jill are racing their toboggans down an icy hill. Jack and Jill are of similar mass and are using the same type of toboggan. When Jack is a certain distance from the end of the race they are travelling with the same velocity. Jack is behind Jill and decides that if he is going to win the race he must lighten his toboggan, so he pushes a box containing their ice-skating gear off the side of his toboggan. Question 11 Explain, giving clear reasons, whether this will be a successful way for Jack to catch up to Jill and help him win the race. This question is worth 4 marks, so you should give at least four distinct answers. You must also answer the question in a clear manner. It is always useful to put your answer in point form. This question was meant to be a momentum question, but a lot of students used energy and forces to try to explain the answer. It was much simpler as a momentum question. If Jack pushes the box off the side of his toboggan, then it will have gained sideways momentum. As momentum is always conserved, then Jack also gains sideways momentum, but in the opposite direction. The box will still have its original downward momentum, so it will continue down next to Jack From a conservation of momentum, there will be no change in the forward momentum of the box or toboggan Lightening his toboggan will not cause Jack to gain or lose downhill momentum, so his speed will stay the same. If Jack really wanted to win the race, then he needed to project the box backwards Jill will remain in front and win the race. Page 4

5 000 The standing 400 m time for a car is the time that it takes to travel 400 m on a level road, accelerating from rest. The standing 400 m time of a car was 16.0 s. Question 1 Calculate the acceleration of the car, assuming constant acceleration for the entire journey. Q x ut at a 16 a = 3.1 ms - 16 The acceleration of the car was calculated from the equations for uniform acceleration to be 3.1 m s -. The most common error here was first to calculate the average velocity and then use this value in one of the equations for uniform acceleration. This was a remarkably common error. Question Assuming constant acceleration, calculate the speed of the car at the end of 400 m. 1 1 x vt at 400 v v = 50 m/s 16 The speed of the car was calculated via the equations for uniform acceleration, resulting in 50 m s -1. There were a number of correct consequential answers for students who obtained the incorrect answer to Question 1. This question was usually correctly answered, either directly or consequentially. The test on the car was repeated in the opposite direction and the standing 400 m time was 18.0 s. The momentum of the car at the end of the first 400 m may be represented in magnitude and direction by the vector shown below. Page 5

6 Question 3 Which one of the vectors (A G) best represents the momentum change of the car, between the end of the first and the end of the second run? This question is testing your understanding of vectors using momentum. The cars initial momentum was to the right. The cars final momentum was to the left. The final momentum minus the initial momentum gives a change in momentum. expected to do this with the vectors. Thus: - This minus just changes the direction of the vector that follows it. Therefore you get: Which gives you a change in momentum like this. So the answer is D. p p f p. You are i The change in momentum is defined as final momentum initial momentum. Subtracting the initial momentum vector from the final momentum vector resulted in vector D as the change in momentum vector In a car the driver s head is moving horizontally at 8.0 m s -1 and collides with an air bag as illustrated in Figure 1. The time taken for the driver s head to come to a complete stop is s. This collision may be modelled as a simple horizontal collision between the head of mass 7.0 kg and the air bag. Question 1 Calculate the magnitude of the average contact force that the air bag exerts on the driver s head during this collision. Page 6

7 350N Use F t = m v. F = = 350 N The average force in the collision was 350 N for the driver s head in context A and 4000 N for the footballer in context B. The average mark for this question was 3.1/4, indicating that students had a solid understanding if the impulse-momentum concept. Most students attempted this question. Question Explain why the driver is less likely to suffer head injury in a collision with the air bag than if his head collided with the car dashboard, or other hard surface. The airbag is designed to increase the time of the collision. It expands rapidly and is already deflating by the time the head comes into contact with it. This deflating bag increases the time of collision greatly. From the Impulse equation F t = m v it can be seen that an increase in t for a fixed value of m v will lead to a decrease in F. Hard surfaces result in shorter contact times, or softer surfaces result in longer contact times The larger F is, the greater the risk that parts of the body will undergo forces that will push the body beyond its elastic limit, resulting in injury. The explanation required students to cover these main points. Impulse equals change in momentum. ( Ft = mv mu) For this collision the change in momentum is a fixed quantity. Hard surfaces result in shorter contact times, or softer surfaces result in longer contact times. Hence, with the change in momentum fixed, then shorter contact times result in larger forces, or longer contact times result in smaller forces. The above could also be addressed via a work-energy or force-acceleration approach as well. The average mark for this question was.1/4, a disappointing result for what was expected to be a fairly straightforward question. The major problem encountered by students was in failing to mention the significance of the fixed aspect of the change in momentum. Instead, these students simply gave bald statements such as longer time results in a smaller force, which was insufficient to score full marks for this question. Page 7